Welcome to Calculus. I'm Professor Ghrist. We're about to begin Lecture
33 on complex volumes. Computing volumes in general is not so
easy a thing to do. In fact, you're going to spend
a significant amount of time in multivariable calculus
worrying over such problems. But as we'll see in this lesson,
some volume problems are solvable if you put
the right spin on it. The types of objects whose volumes we
can compute are necessarily limited. Most solids are too complex. However, there is a class of solids for which our single variable
calculus will work. These are the volumes of revolution, that is solids obtained by rotating
some two dimensional shape about an axis,
sweeping out a three dimensional volume. There are two principle ways
to decompose such an object into simple volume elements. The first is by taking
slices orthogonal to the axis of revolution and
progressing along the axis. This tends to lead to
disc-like volume elements. Of course, the other way is
parallel to the axis of revolution. Slicing about a region that is parallel leads to a cylindrical
volume element in general. Let's do some examples to
see these two approaches.. Let us consider this solid formed
by rotating a disk about an axis. One might describe this
object as doughnut like. What's the volume of such an object? Let's set up coordinates so
that the Y axis is the axis of revolution. And that the disc that is rotated
there about is of radius a. And the center of the disc is located a distance capital R away
from the axis of rotation. Then, decomposing this volume
by slices that are parallel to the axis of revolution,
gives us a cylindrical volume element. Let's compute that volume element and
then integrate. The volume element intersects
this disk along a vertical strip. What is the height of that
vertical strip if we use x to denote our distance
from the axis of rotation? Then by building the appropriate
right triangle and recalling that this disc has radius a,
we see that the height of this cylinder is twice the square
root of a squared- quantity (x-R) quantity squared. Thus the volume element,
being cylindrical, is what? We have to take the circumference,
that is 2 pi x, and multiply it by the height,
twice root a squared- (x-R) squared, and then multiply that by the thickness,
dx. We integrate this to get the volume. And now we have the integral of 4 pi x, square root of a squared-
(x-R) squared dx. What are the limits on x? x is going from capital R-a to R+a. This integral is not trivial, but it's easier if we perform
a change of coordinates. Let u be x-R. We're centering the coordinates
at the origin of this disc. Then we're integrating,
as u goes from -a to a, 4 pi times x, which is u + r, times square root of a squared- u squared,
du. If we distribute that multiplication and
split it up in to two integrals, then our job becomes easier. Because we note that the integral
on the left has an odd integrand, that is an odd function in u that is
integrated over a symmetric domain. Therefore, the former integral vanishes,
and goes to 0, and the only thing we have to
compute is that second Integral. Now I'm going to break
that up a little bit. I'm going to factor the 2 pi R. What's left over is twice the square
root of a squared- u squared du. Why have I done that? Because if I pull out that constant,
2 pi R, then what is left over is
an integral that I recognize. It is an integral which
computes the area of that disc, which we certainly know
to be pi a squared, giving a final volume of
2 pi squared R a squared. If we repeat this computation using
a volume element obtained by slicing orthogonal to the axis revolution,
then what will we obtain? This volume element is going
to be an annular region consisting of some large thickened disk
minus a smaller concentric thickened disk. We need to determine the radii
of this outer and inner disk. To do so, let's construct, again, some right triangles in
the cross-sectional disk. And knowing that these right
triangles have hypotenuse, a, and height y, then we see that the width is square root of a squared- y squared. That means what? Well the volume element is going to be
the volume of the thickened outer disc, that is pi times quantity (R + square root of a squared- y squared) squared. We have to subtract the volume of the
inner disc, that is the disc with radius capital R- square root
of a squared- y squared. And this volume element is,
of course, multiplied by dy and then algebraically simplified. We see that we have identical
terms in our squares and we're subtracting
the latter from the former. So by expanding and collecting terms, we see we obtain 2 pi R square
root of a squared- y squared minus negative 2 pi R square
root of a squared- y squared. All of that times dy. Thus, we can compute the volume as
the integral of this volume element. As the integral of 4 pi R square,
square root of a squared- y squared dy. And we see that although we're using
a different volume element and a different variable,
that integrating this as y goes from -a to a yields the same integral
as we saw previously. It is 2 pi R times the integral
that gives you the area of this disc, leading to,
again, the final answer of 2 pi squared R a squared. Now you should perhaps be thinking
at this point that it seems as though this volume is really the area
of the cross sectional disc times the circumference traced out
by the center of that disc. But we haven't proven that that's true,
but keep it in mind. Let's move on to a different example. In this one compute the volume of
a ball with a cylindrical hole drilled through it, like a bead. Let's say that we set the y-axis
to be the axis of revolution, and that our ball has radius capital R and
the hole drilled has radius a. If we choose slices that are parallel
to the axis of revolution, then we have a cylindrical volume element,
a cylinder with radius x that varies. In this case,
the volume element is 2 pi x, the circumference of that cylinder times
the height, which is twice square root of R squared- x squared dx. Integrating this to obtain
the volume gives, what? The integral of 4 pi x square
root of R squared- x squared dx, as x goes from a to R. Pay very careful attention
to those limits. Make sure you understand why that is true. Then doing this integral won't be so bad with a new substitution,
u = R squared- x squared. That leads to the integral
of 2 pi times the integral as u goes from h-squared to 0, where h is the half height of this object,
if you will. We're integrating negative
square root of u, du. That leads to a final answer
of- 2 pi times the integral of the square root of u. That's two-thirds u to the three-halves. Evaluating as u goes from
h squared to 0 gives what? Four-thirds pi h cubed. We can also compute this volume
by choosing a volume element that is orthogonal to
the axis of revolution. In this case, those slices at some
constant y are annular regions. And to compute the volume element,
we need to determine the outer radius and the inner radius of these
annuli as a function of y. Well in this case, the outer radius is, through a right triangle computation,
square root of R squared- y squared. The associated area of that disc
is pi times that quantity squared. When we subtract off the area of the inner
radius disc, that's removing pi a squared. This times the thickness,
dy, gives the volume of it. We can simplify that a little bit using
what we know about this height, h. This yields the volume element, pi times quantity (h
squared- y squared) dy. And thus computing the volume as
the integral of the volume element yields pi times the integral of h squared-
y squared dy, as y goes from -h to h. This is a simple integral yielding
the same answer as before, four-thirds pi h cubed. Remarkable in that a and
r are not explicitly stated. Let's do one last example. Compute the volume of a vase
formed by rotating the curve x = square root of sine y + y over pi, about the y axis, y goes from 0 to 2 pi. Using an orthogonal slice yields
discs as the volume element, discs of radius x. Thus the volume element
is pi x squared dy. Substituting in our function for x yields pi times quantity
(sine y + y over pi) dy. Integrating this to obtain the volume as y goes from 0 to 2 pi gives
a very simple answer. We get when we -pi cosine y + y squared over 2, evaluated from 0 to 2 pi. I'll let you compute that
the volume is 2 pi squared. Now when you're doing homework
problems on volumes of revolution, there are couple of
things to keep in mind. The first is that there's no one formula that will do all volume computations for
you. You're going to have to think about
how to decompose your region. Always think in terms of whether
you're slicing parallel to or orthogonal to the axis of revolution. And that will inform what you use for
the associated volume element. This ends our introduction
to volume in this dimension. In our next lesson,
we're gong to be a bit more ambitious. We're going to ascend
to higher dimensions, introduce higher dimensional shapes and
volumes. And show how calculus can
make sense of the geometry of the fourth dimension and beyond.